After cutting out the individual chromosomes, our group measured the
length of the long and short arms of each chromosome and plotted them
against each other. This was to help establish a size relationship so
that we could determine which chromosomes were sister chromatids.
With that information, we were able to put together the following
karyotype. As you will notice, all the chromosomes, except one, have
three sets. Effectively, it is a triploid cell (with that one exception).
We will discuss two scenarios that could have led to such a karyotype.
1. Parent 1 would supply a diploid gamete and Parent 2 would supply a
haploid gamete with a single missing chromosome (the result of a
non-disjunction event during Meiosis I or II). The diploid gamete also
obviously resulted from complete non-disjunction, in which one daughter
cell received all the chromatids and the other received none. So, each
gamete in this case should have had 7 chromosomes, making the normal 14
diploid cell. However, the diploid parent (#1) provided 14 chromosomes
and the haploid parent (#2) provided only 6, making the 20 total that is
observed. Thus, a triploid is produced (minus that one set).
2. In the second scenario, Parent 1 would again be a diploid gamete
resulting from non-disjunction. However, 1 of the sister chromatids did
separate correctly, leaving the gamete cell with 13 chromosomes. Parent 2
would then be normal haploid gamete with 7 chromosomes. This total again
adds to the 20 chromosomes that are observed.
length of the long and short arms of each chromosome and plotted them
against each other. This was to help establish a size relationship so
that we could determine which chromosomes were sister chromatids.
With that information, we were able to put together the following
karyotype. As you will notice, all the chromosomes, except one, have
three sets. Effectively, it is a triploid cell (with that one exception).
We will discuss two scenarios that could have led to such a karyotype.
1. Parent 1 would supply a diploid gamete and Parent 2 would supply a
haploid gamete with a single missing chromosome (the result of a
non-disjunction event during Meiosis I or II). The diploid gamete also
obviously resulted from complete non-disjunction, in which one daughter
cell received all the chromatids and the other received none. So, each
gamete in this case should have had 7 chromosomes, making the normal 14
diploid cell. However, the diploid parent (#1) provided 14 chromosomes
and the haploid parent (#2) provided only 6, making the 20 total that is
observed. Thus, a triploid is produced (minus that one set).
2. In the second scenario, Parent 1 would again be a diploid gamete
resulting from non-disjunction. However, 1 of the sister chromatids did
separate correctly, leaving the gamete cell with 13 chromosomes. Parent 2
would then be normal haploid gamete with 7 chromosomes. This total again
adds to the 20 chromosomes that are observed.